the_encyclopedia_of_examinations_in_hong_kongwikiaorg_zh-20200214-history
M2 DSE 2013
Q1 : \frac{d}{dx}(sin2x) : =\lim_{\Delta x\to 0} \frac{2sin\frac{[2(x+\Delta x)-2x]}{2}cos\frac{[2(x+\Delta x)+2x]}{2}}{\Delta x} : =\lim_{\Delta x\to 0} 2\frac{sin\Delta x}{\Delta x}cos(2x+2\Delta x) : =2\cdot(1)\cdot cos[2x+2\cdot(0)] : =2cos2x Q2 : (1+ax)^n : =1+nax+\frac{n(n-1)}{2}a^2x^2+... The coefficients of x and x^2 are 20 and 180 respectively. So we have : \begin{cases} na=-20 & \\ \frac{n(n-1)}{2}a^2=180 \end{cases} On solving, we have : \begin{cases} a=-2 \\ n=10 \end{cases} Q3 Let P(n) be the proposition " 1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{4n+1}{3n+1} for all positive integers n ." For n=1 , L.S. =1+\frac{1}{1\times4}=\frac{5}{4} R.S. =\frac{4\cdot(1)+1}{3\cdot(1)+1}=\frac{5}{4} L.S.=R.S. \Rightarrow P(1) is true. Next, suppose P(n) is true for some positive integers k , i.e. " 1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{4k+1}{3k+1} " Then for n=k+1 , L.S. =1+\frac{1}{1\times 4}+\frac{1}{4\times 7}+\frac{1}{7\times 10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{[3(k+1)-2][3(k+1)+1]} : =\frac{4k+1}{3k+1}+\frac{1}{(3k+1)(3k+4)} : =\frac{4k+1}{3k+1}+\frac{1}{(3k+1)(3k+4)} : =\frac{(4k+1)(3k+4)+1}{(3k+1)(3k+4)} : =\frac{12k^2+19k+5}{(3k+1)(3k+4)} : =\frac{(3k+1)(4k+5)}{(3k+1)(3k+4)} : =\frac{4k+5}{3k+4} R.S. =\frac{4(k+1)+1}{3(k+1)+1} : =\frac{4k+5}{3k+4} L.S.=R.S. \Rightarrow P(k+1) is true. By the principle of mathematical induction, P(n) is true for all positive integers n . Q4 a) : \frac{dy}{dx}=e^x-1 : \int \frac{dy}{dx} dx=\int e^x-1 dx : y=e^x-x+C At (1,e) , : (e)=e^{(1)}-(1)+C \Rightarrow C=1 Therefore y=e^x-x+1 is the required equation. b) At x=0 , : y=e^{(0)}-(0)+1=2 : \frac{dy}{dx}|_{x=0}=e^{(0)}-1=0 So the slope of the required tangent has a slope of 0 and an y-intercept of 2. Therefore y=2 is the required tangent. Q5 a) Since the sign of f'(x) changes from + to - as x passes through 0 in the positive direction, a maximum point occurs at x=0 . Since the sign of f''(x) changes as x passes through 1 and -1, inflexion points occur at x=1 and x=-1 . At x=0 , : f(0)=\frac{3-3\cdot(0)^2}{3+(0)^2}=1 At x=1 , : f(1)=\frac{3-3\cdot(1)^2}{3+(1)^2}=0 At x=-1 , : f(-1)=\frac{3-3\cdot(-1)^2}{3+(-1)^2}=0 So a maximum point occurs at (0,1) and two inflexion points occur at (1,0) and (-1,0) . b) Since x^2\geq 0 for any real number x , x^2+3\geq 3 for any real number x . f(x) has no vertical asymptote. Since f(x)=\frac{3-3x^2}{3+3x^2}=\frac{-3(x^2+3)+12}{3+x^2}=-3+\frac{12}{3+x^2} , : \lim_{x\to \pm\infty} f(x)=\lim_{x\to \pm\infty} -3+\frac{12}{3+x^2}=-3+(0)=-3 . So y=-3 is a horizontal asymptote of f(x) . No oblique asymptote. c) To be drawn Q6 a) Required area : =\int_0^4 -\frac{x^2}{2}+2x+4-4 dx+\int_4^5 4-(-\frac{x^2}{2}+2x+4-4) dx : =-\frac{x^3}{6}|_0^4+x^2|_0^4+\frac{x^3}{6}|_4^5-x^2|_4^5 : =-\frac{64}{6}+16+\frac{125}{6}-\frac{64}{6}-25+16 : =-\frac{13}{2} sq. units b) Required volume : =|\pi \int_0^4 (-\frac{x^2}{2}+2x+4-4)^2 dx+\pi \int_4^5 [4-(-\frac{x^2}{2}+2x+4)]^2 dx| : =|\pi \int_0^4 (-\frac{x^2}{2}+2x)^2 dx+\pi \int_4^5 (\frac{x^2}{2}-2x)]^2 dx| : =|\pi \int_0^5 (-\frac{x^2}{2}+2x)^2 dx| : =|\pi \int_0^5 \frac{x^4}{4}-2x^3+4x^2 dx| : =|\pi (\frac{x^5}{20}-\frac{x^4}{2}+\frac{x^3}{4})|_0^5| : =\frac{123\pi}{12} cu. units Q7 a) : tanx : =\frac{sinx}{cosx} : =\frac{sinx}{cosx}\times \frac{2cosx}{2cosx} : =\frac{2sinxcosx}{2cos^2x} : =\frac{sin2x}{1+cos2x} b) : tany : =\frac{sin2y}{1+cos2y} (By (a)) : =\frac{sin2y}{1+cos2y}\times \frac{sec2y}{sec2y} : =\frac{tan2y}{sec2y+1} : =\frac{\frac{sin4y}{1+cos4y}}{sec2y+1} (By (a)) : =\frac{sin4y}{(1+cos4y)(sec2y+1)}\times \frac{sec4y}{sec4y} : =\frac{tan4y}{(sec4y+1)(sec2y+1)} : =\frac{\frac{sin8y}{1+cos8y}}{(sec4y+1)(sec2y+1)} (By (a)) : =\frac{sin8y}{(1+cos8y)(sec4y+1)(sec2y+1)}\times \frac{cos2ycos4y}{cos2ycos4y} : =\frac{sin8ycos2ycos4y}{(1+cos8y)(1+cos4y)(1+cos2y)} Q8 : M=\begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 1 \\ k & 0 & 0 \end{bmatrix}, k \ne 0 a) : detM= : 1 \begin{vmatrix} 1 & 1 \\ 0 & 0 \end{vmatrix} - k \begin{vmatrix} 0 & 1 \\ k & 0 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ k & 0 \end{vmatrix} : =k^2 : M^{-1} : =\frac{1}{detM}{\begin{bmatrix} 0 & k & -k \\ 0 & 0 & k^2 \\ k & -1 & 1 \end{bmatrix}}^T : =\frac{1}{(k^2)}\begin{bmatrix} 0 & 0 & k \\ k & 0 & -1 \\ -k & k^2 & 1 \end{bmatrix} b) : M\begin{bmatrix} x \\ 1 \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} : \begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 1 \\ k & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ 1 \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 2 \\ 1 \end{bmatrix} : \begin{cases} x+k=2 ......(1) & \\ 1+z=2 ......(2) & \\ kx=1 ......(3) \end{cases} On solving (1) and (3), we have k=1 . Q9 a) : \left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 2 & 1-2a & 2-b & a+4 \\ 3 & 1-3a & 3-ab & 4 \end{array}\right]\sim \left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 0 & 1 & -b & a \\ 0 & 1 & -ab & -2 \end{array}\right] We require the system to have infinitely solutions, i.e. : \begin{cases} a=-2 ......(1) & \\ -b=-ab ......(2) & \end{cases} From (2), a=1 or b=0 , but since we have already required a=-2 , we reject the solution a=1 . Therefore, for infinitely many solutions we are left with : \begin{cases} a=-2 & \\ b=0 & \end{cases} b) Since (E) has infinitely many solutions, a=-2 and b=0 . : \left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 2 & 1-2a & 2-b & a+4 \\ 3 & 1-3a & 3-ab & 4 \end{array}\right]\sim \left[\begin{array}{rrr|r} 1 & -a & 1 & 2 \\ 0 & 1 & -b & a \\ 0 & 1 & -ab & -2 \end{array}\right] (From (a)) : \sim \left[\begin{array}{rrr|r} 1 & -(-2) & 1 & 2 \\ 0 & 1 & -(0) & (-2) \\ 0 & 1 & -(-2)(0) & -2 \end{array}\right]\sim \left[\begin{array}{rrr|r} 1 & 2 & 1 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 1 & 0 & -2 \end{array}\right] : \sim \left[\begin{array}{rrr|r} 1 & 2 & 1 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right]\Rightarrow \begin{cases} z=t & \\ y=-2-t & \\ x=2-t-2(-2-t)=6+t & \end{cases} for any real number t. Q10 a) : \overrightarrow{ON} : =\frac{k}{k+1}\overrightarrow{OA}+\frac{1}{k+1}\overrightarrow{OB} : =\frac{k}{k+1}(2\hat{\imath})+\frac{1}{k+1}(\hat{\imath}+2\hat{\jmath}) : =\frac{2k+1}{k+1}\hat{\imath}+\frac{2}{k+1}\hat{\jmath} b) : \overrightarrow{MB} : =\overrightarrow{OB}-\overrightarrow{OM} : =\overrightarrow{OB}-\frac{1}{2}\overrightarrow{OA} : =(\hat{\imath}+2\hat{\jmath})-[\frac{1}{2}(2\hat{\imath})] : =2\hat{\jmath} : \overrightarrow{OA}\cdot\overrightarrow{MB}=0 since \hat{\imath}\perp\hat{\jmath} : \Rightarrow \angle PMA=90^{\circ} : \angle ANO : =180^{\circ}-\angle PMA (opp. \angle , cyclic quad.) : =180^{\circ}-(90^{\circ})=90^{\circ} : \Rightarrow \overrightarrow{ON}\perp\overrightarrow{AB}\Rightarrow \overrightarrow{ON}\cdot\overrightarrow{AB}=0 : \overrightarrow{AB} : =\overrightarrow{OB}-\overrightarrow{OA} : =(\hat{\imath}+2\hat{\jmath})-(2\hat{\imath}) : =-\hat{\imath}+2\hat{\jmath} : \overrightarrow{ON}\cdot\overrightarrow{AB}=0 : (\frac{2k+1}{k+1}\hat{\imath}+\frac{2}{k+1}\hat{\jmath})\cdot(-\hat{\imath}+2\hat{\jmath})=0 : -\frac{2k+1}{k+1}+\frac{4}{k+1}=0 : 3-2k=0 : k=\frac{3}{2} Q11 a) : \frac{d}{d\theta}[\ln(sec\theta+tan\theta)] : =\frac{1}{sec\theta+tan\theta}sec\theta tan\theta+sec^2\theta : =\frac{sec\theta+tan\theta}{sec\theta+tan\theta}sec\theta : =sec\theta Therefore \int sec\theta d\theta=\int\frac{d}{d\theta}\ln(sec\theta+tan\theta)d\theta+C=\ln(sec\theta+tan\theta)+C b) i) : \int \frac{du}{\sqrt{u^2-1}} Let u=sec\theta for 0\le\theta\le\frac{\pi}{2} , so that du=sec\theta tan\theta d\theta . : \int \frac{du}{\sqrt{u^2-1}} : =\int \frac{sec\theta tan\theta d\theta}{\sqrt{sec^2\theta-1}} : =\int \frac{sec\theta tan\theta d\theta}{\sqrt{tan^2\theta}} : =\int sec\theta : =\ln(sec\theta+tan\theta)+C (By (a)) : =\ln(u+\sqrt{u^2-1})+C, u\ge 1 Diagram to be inserted here ii) : \int_0^1 \frac{2x}{\sqrt{4x^2+x^4+3}} dx : =\int_0^1 \frac{2x}{\sqrt{x^4+4x^2+4-1}} dx : =\int_0^1 \frac{d(x^2+2)}{\sqrt{(x^2+2)^2-1}} : =\ln[(x^2+2)+\sqrt{(x^2+2)^2-1}]|_0^1 (By (b)) : =\ln[\frac{((1)^2+2)+\sqrt{((1)^2+2)^2-1}}{((0)^2+2)+\sqrt{((0)^2+2)^2-1}}] : =\ln(\frac{3+2\sqrt{2}}{2+\sqrt{3}}) : =\ln(\frac{3+2\sqrt{2}}{2+\sqrt{3}}\frac{2-sqrt{3}}{2-sqrt{3}}) : =\ln[\frac{(3+2\sqrt{2})(2-sqrt{3})}{4-3}] : =\ln(4\sqrt{2}-3\sqrt{3}-2\sqrt{6}+6) c) : t=tan\phi : 1=sec^2\phi\frac{d\phi}{dt} : \frac{d\phi}{dt}=cos^2\phi : \frac{d\phi}{dt}=(\frac{1}{\sqrt{1+t^2}})^2=\frac{1}{1+t^2} So d\phi=\frac{dt}{1+t^2} : \int_0^{\frac{\pi}{4}} \frac{tan\phi}{\sqrt{1+2cos^2\phi}} d\phi : =\int_0^1 \frac{t}{\sqrt{1+\frac{2}{1+t^2}}} (\frac{dt}{1+t^2}) : =\int_0^1 \frac{t dt}{\sqrt{(1+t^2)^2+2(1+t^2)}} : =\int_0^1 \frac{t dt}{\sqrt{(1+t^2)^2+2(1+t^2)+1-1}} : =\int_0^1 \frac{t dt}{\sqrt{[(1+t^2)+1]^2-1}} : =\frac{1}{2}\int_0^1 \frac{d(t^2+2)}{\sqrt{(t^2+2)^2-1}} : =\frac{1}{2}\ln(4\sqrt{2}-3\sqrt{3}-2\sqrt{6}+6) (By (b)(ii)) Q12 a) i) : T : =\frac{x}{7}+\frac{\sqrt{(40-x)^2+30^2}}{1.4} : =\frac{x+5\sqrt{(40-x)^2+900}}{7} ii) : \frac{dT}{dx}=\frac{1+5\cdot\frac{1}{2}[(40-x)^2+900]^{-\frac{1}{2}}[2\cdot(40-x)\cdot(-1)]}{7} When T is minimum, \frac{dT}{dx}=0 . So, : (0)=1+5\cdot\frac{1}{2}[(40-x)^2+900]^{-\frac{1}{2}}[2\cdot(40-x)\cdot(-1)] : 0=1-5[(40-x)^2+900]^{-\frac{1}{2}}(40-x)] : 1=5[(40-x)^2+900]^{-\frac{1}{2}}(40-x)] : (40-x)^2+900=[5(40-x)]^2 : [(40-x)^2+900]=25(40-x)^2 : 24(40-x)^2-900=0 : (40-x)^2-\frac{900}{24}=0 : 2x^2-160x+3125=0 : x=\frac{160}{2\cdot2}\pm\frac{\sqrt{(-160)^2-4\cdot(2)\cdot(3125)}}{2\cdot2} : x=40\pm\frac{75}{2}=40\pm5\frac{3}{2} First derivative test hand-drawn table here So, x=40+5\sqrt{\frac{3}{2}} for minimum T . : QB : =\sqrt{[40-(40+5\sqrt{\frac{3}{2}})]^2+30^2} (Pyth. thm.) : =\sqrt{\frac{1875}{2}} : =25\sqrt{\frac{3}{2}}\frac{\sqrt{2}}{\sqrt{2}} : =\frac{25\sqrt{6}}{2} m b) i) Q13 a) : M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}, tr(M)\equiv a+d. : BAB^{-1}=\begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix} i) : MN=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} e & f \\ g & h \end{bmatrix}=\begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix} : NM=\begin{bmatrix} e & f \\ g & h \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} ae+fc & eb+fd \\ ga+ch & gb+hd \end{bmatrix} : tr(MN)=ae+bg+cf+dh : tr(NM)=ae+fc+gb+hd=tr(MN) ii) : tr(BAB^{-1}) : =tr(AB^{-1}B) (By (a)(i)) : =tr(AI) : =tr(A) Since tr(BAB^{-1})=1+3=4 , tr(A)=4. iii) : det(BAB^{-1}) : =detBdetAdet(B^{-1}) : =detBdet(B^{-1})detA : =det(BB^{-1})detA : =det(I)detA : =(1)\cdot detA : =detA Since det(BAB^{-1})=1\times3-0\times0=3 , detA=3. b) : C=\begin{bmatrix} p & q \\ r & s \end{bmatrix} : C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda_1\begin{bmatrix} x \\ y \end{bmatrix} and C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda_2\begin{bmatrix} x \\ y \end{bmatrix} for \begin{bmatrix} x \\ y \end{bmatrix}\ne0 , \lambda_1\ne\lambda_2 . i) : C\begin{bmatrix} x \\ y \end{bmatrix}=\lambda_i\begin{bmatrix} x \\ y \end{bmatrix} for i=1, 2 : (C-\lambda_i)\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix} : \begin{bmatrix} p-lambda_i & q \\ r & s-lambda_i \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix} As we have required \begin{bmatrix} x \\ y \end{bmatrix} to be non-trivial, i.e. \begin{bmatrix} x \\ y \end{bmatrix}\ne0 , we must have : \begin{vmatrix} p-\lambda_1 & q \\ r & s-\lambda_1 \end{vmatrix}=\begin{vmatrix} p-\lambda_2 & q \\ r & s-\lambda_2 \end{vmatrix}=0 . ii) : \begin{vmatrix} p-\lambda_i & q \\ r & s-\lambda_i \end{vmatrix}=0 for i=1, 2 : (p-\lambda_i)(s-\lambda_i)-qr=0 : ps-p\lambda_i-s\lambda_i+\lambda_i^2-qr=0 : \lambda_i^2-(p+s)\lambda_i+(ps-qr)=0 Notice that detC=ps-qr and trC=p+s . So, \lambda_i^2-trC\lambda_i+detC=0 for i=1, 2 . \lambda_1 and \lambda_2 are therefore the roots of the equation \lambda^2-trC\lambda+detC=0 . c) : \lambda^2-trC\lambda+detC=0 (By (b)(ii)) : \lambda^2-(4)\lambda+(3)=0 (By (a)(ii) and (iii)) : (\lambda-1)(\lambda-3)=0 : \lambda=1 or 3 Q14